Question

In the circuit shown below the average power consumed by the 1 $\Omega$ resistor is

• A. 50 W
• B. 1050 W
• C. 5000 W
• D. 10100 W

Answer 1

The correct option is B 1050 W



Current ,

$I=\frac{100\sqrt{2}cos3000t}{1+j{\omega }_{1}L}+\frac{10\sqrt{2}sin1000t}{1+j{\omega }_{2}L}$

$I=\frac{100\sqrt{2}cos3000t}{1+j\times 3000\times {10}^{-3}}+\frac{10\sqrt{2}sin1000t}{1+j\times 1000\times {10}^{-3}}$

=$\frac{100\sqrt{2}cos3000t}{1+j3}+\frac{10\sqrt{2}sin1000t}{1+j1}$

$=\frac{100\sqrt{2}cos3000t}{\sqrt{10}\angle {\varphi }_1}+\frac{10sin\left(1000t\right)\times \sqrt{2}}{\sqrt{2}\angle {\varphi }_1}$

where,

${\varphi }_1=ta{n}^{-1}\left(3\right)and{\varphi }_2=ta{n}^{-1}\left(1\right)$

So, $I=\frac{100\sqrt{2}}{\sqrt{10}}cos(3000t-{\varphi }_1)+\frac{10sin\left(1000t\right)\times \sqrt{2}}{\sqrt{2}\angle {\varphi }_2}$

$\Rightarrow So,RMSofIis$

$I_{rms}=\sqrt{\frac{1}{2}[\frac{1000\times 2}{10}+100]}=\sqrt{1050}$

So, power dissipated is

P = $I_{rms}^2\times R=1050\times 1=1050Watt$